\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{1}{3}\sqrt{2x}-2\sqrt{2x}+3\sqrt{2x}=12\\ \Leftrightarrow\dfrac{4}{3}\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=9\\ \Leftrightarrow2x=81\Leftrightarrow x=\dfrac{81}{2}\left(tm\right)\)

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x+3\right)^2}=12\)

=>\(\left|x+3\right|=12\)

=>\(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)

b: ĐKXĐ: x>=1

\(\sqrt{25x-25}-\sqrt{9x-9}=10\)

=>\(5\sqrt{x-1}-3\sqrt{x-1}=10\)

=>\(2\sqrt{x-1}=10\)

=>x-1=25

=>x=26(nhận)

\(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\left(x\ge1\right)\)

\(< =>5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

\(< =>30\sqrt{x-1}-15\sqrt{x-1}=36+6\sqrt{x-1}\)

\(< =>9\sqrt{x-1}=36\\ < =>\sqrt{x-1}=4\\ < =>x-1=16\\ < =>x=17\left(tm\right)\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{1}{3}\sqrt{x-1}-\sqrt{x-1}=6\)

=>\(1.5\cdot\sqrt{x-1}=6\)

=>\(\sqrt{x-1}=4\)

=>x-1=16

=>x=17

\(a,\sqrt{25x^2}=10\)

\(\sqrt{\left(5x\right)^2}=10\)

\(5x=10\)

\(x=2\)

b. <=> \(\sqrt{4\left(x^2-1\right)}=2\sqrt{15}\)     ĐKXĐ: x>=1,x>=-1

<=> \(4\left(x^2-1\right)=60\Leftrightarrow x^2-1=15\Leftrightarrow x^2-16=0\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)

<=>x=+-4

câu b:

chuyển 2 căn 15 sang vế phải , 

sau đó bình phương cả 2 vế lên thì ta mất dấu căn ở căn (4(x^2-1)) từ đó giải ra ptr như bình thường , ra 2 nghiệm là 4 và -4 

câu c :

bình phương cả hai vế lên , vế trái dùng hằng đẳng thức phân tích dần ra là đc 

câu d 

chuyển  căn x-5 sang vế phải rồi bình phương hai vế, mất dấu căn thì giải ptrinh như bth thôi 

:P lười làm full bài 

a/ \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐKXĐ : \(x\ge1\))

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow2\sqrt{x-1}=2\Leftrightarrow x-1=1\Leftrightarrow x=2\)

b/ \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)

\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}+3=0\)

<=> 3 = 0 (vô lý)

=> pt vô nghiệm.

c/ \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (ĐKXĐ : x>-5/7)

\(\Leftrightarrow9x-7=7x+5\Leftrightarrow2x=12\Leftrightarrow x=6\)

d/ \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\) (ĐKXĐ : \(x\ge\frac{3}{2}\))

\(\Leftrightarrow2x-3=4\left(x-1\Leftrightarrow\right)2x=1\Leftrightarrow x=\frac{1}{2}\) (loại)

Vậy pt vô nghiệm.

6) \(\sqrt{x^2-4x+1}=x\left(x\ge0\right)\) 

\(\Leftrightarrow x^2-4x+1=x^2\)

\(\Leftrightarrow x^2-x^2=4x-1\)

\(\Leftrightarrow4x=1\)

\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\) 

8) \(\sqrt{x^2-x-6}=\sqrt{x-3}\left(x\ge3\right)\) 

\(\Leftrightarrow x^2-x-6=x-3\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

9) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\left(x\ge1\right)\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=1+1\)

\(\Leftrightarrow x=2\left(tm\right)\)

Đề bài là Rút gọn biểu thức nha . Mình quên ghi ^^

\(A=\sqrt{80}+\sqrt{45}+\sqrt{5}=\sqrt{16.5}+\sqrt{9.5}+\sqrt{5}\)

\(=4\sqrt{5}+3\sqrt{5}+\sqrt{5}=8\sqrt{5}\)

\(B=\frac{5}{\sqrt{10}}+3,5\sqrt{40}=\sqrt{\frac{25}{10}}+3,5\sqrt{16.2,5}\)

\(=\sqrt{2,5}+3,5.4\sqrt{2,5}=15\sqrt{2,5}\)

\(C=\frac{1}{\sqrt{3}-2}+\frac{\sqrt{300}}{10}-\sqrt{12}\)

\(=\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{100.3}}{10}-\sqrt{4.3}\)

\(=-\sqrt{3}-2+\sqrt{3}-2\sqrt{3}=-2\sqrt{3}-2\)

\(D=4\sqrt{x}+2\sqrt{x^2}-\sqrt{16x}=4\sqrt{x}+2x-4\sqrt{x}=2x\) ( do \(x\ge0\))

\(F=\frac{a-2\sqrt{a}}{\sqrt{a}-2}=\frac{\sqrt{a}.\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=\sqrt{a}\)

mk chỉnh đề

\(E=\sqrt{25x+25}-\sqrt{9x+9}+\sqrt{4x+4}\)

\(=\sqrt{25\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}\)

\(=5\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=4\sqrt{x+1}\)

\(G=\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{2}{\sqrt{5}-\sqrt{7}}=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}-\frac{2\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)

\(=\sqrt{3}-\sqrt{5}-\sqrt{5}-\sqrt{7}=\sqrt{3}-\sqrt{7}\)

a)

\(\sqrt{4x-4}-\sqrt{9x-9}+\sqrt{25x-25}=4+\sqrt{16x-16}\\ \Leftrightarrow2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}+5\sqrt{x-1}=4\\ \Leftrightarrow0\sqrt{x-1}=4\\ \Rightarrow kh\text{ô}ng\:c\text{ó}\:gi\text{á}\:tr\text{ị}\:x\:th\text{õa}\:m\text{ãn}\)

b)

\(•\sqrt{7-x}+\sqrt{x-5}\le\sqrt{2.\left(7-x+x-5\right)}=2\\ •x^2-12x+38=\left(x-6\right)^2+2\ge2\)

ta thấy \(VT\le2\:v\text{à}\:VP\ge2\) nên \(VT=VP=2\)

đẳng thức xảy ra khi \(\left\{{}\begin{matrix}7-x=x-5\\x-6=0\end{matrix}\right.\Rightarrow x=6\)

vậy nghiệm của phương trình trên là x=6